H3PO4 Stepwise Neutralization

  1. For the triprotic acid, phosphoric acid, H3PO4, there are Ka1, Ka2,
    and Ka3. And we know from the book that the value of Ka3 is too
    small, so we neglect the [H+] contribution from it.

    Normally, we encounter questions involving pH calculations of
    neutralization of a diprotic acid and a base, for example, H2SO4 and
    NaOH. Let say there are 5 mmol of H2SO4 and 10 mmol of NaOH.

    The stepwise reactions are:

    H2SO4 + NaOH ---> NaHSO4 + H2O
    NaHSO4 + NaOH ---> Na2SO4 + H2O

    However, if the reaction is between H3PO4 and NaOH, let say there are
    5 mmol H3PO4 and 15 mmol NaOH.

    The stepwise reactions are:

    H3PO4 + NaOH ---> NaH2PO4 + H2O

    NaH2PO4 + NaOH ---> Na2HPO4 + H2O

    Na2HPO4 + NaOH ---> Na3PO4 + H2O

    So, are the 3 equations above correct? I mean, can we use the 3
    equations above for calculating the pH value, as what we would do for
    the case of the neutralization involving a diprotic acid and NaOH?

    If so, why do we neglect the [H+] from H3PO4?

    Or is it that we ONLY neglect the [H+] from Ka3 when we calculate the
    pH of a solution containing ONLY H3PO4, but we should
    somehow "include" and "consider" the Ka3 when we wish to find the pH
    value of a solution involving neutralization of H3PO4 and NaOH, since
    this neutralization reaction is based on the stoichiometric ratio.

  2. jcsd
  3. You are right.

    You can prove that H3PO4 third dissocitation is negligible as long as H3PO4 is the only source of hydronium ions in solution. ( either generally or u can take a numerical example and try it out yourself . )

    But notice that the reactions u posted :
    H3PO4 + NaOH ---> NaH2PO4 + H2O
    NaH2PO4 + NaOH ---> Na2HPO4 + H2O
    Na2HPO4 + NaOH ---> Na3PO4 + H2O

    are not entirely quantitative. that is , the advancement decreases from one reaction to the other. For each rxn , k = ka/kw . Notice the 3rd rxn is much weaker than the 1st reaction , which is pretty much 'complete'. You can find the ka's and calculate the K of each reaction to convince urself. Finding the pH here is different from that of sulfuric acid.
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